1)

2)

2. ## Re: Simplifying Radicals

What have you tried?

1)
Use basic rules like: $\displaystyle \sqrt[n]{x^{a}}=x^{\frac{a}{n}}$ and $\displaystyle a^{x}\cdot a^{y}=a^{x+y}$
2)
The same over here, you have to know some basic rules like: $\displaystyle (a-b)\cdot (a+b)=a^2-b^2$ and $\displaystyle (a+b)^2=a^2+2ab+b^2$

Try to use them where necessary and simplify.

3. ## Re: Simplifying Radicals

thank you. i will give it another try.

4. ## Re: Simplifying Radicals

Ok! You can show your work afterwards if you want .

5. ## Re: Simplifying Radicals

thanks for the help I did not put the "a" in a common root during the multiplication that was the error : )

6. ## Re: Simplifying Radicals

I have simplified it this way:
1)
$\displaystyle \sqrt[4]{a^{3}}+\sqrt[4]{a}\cdot \sqrt{a}+\frac{3\sqrt[4]{a^{9}}}{\sqrt{a^{3}}}$
$\displaystyle =a^{\frac{3}{4}}+a^{\frac{1}{4}}\cdot a^{\frac{1}{2}}+\frac{3a^{\frac{9}{4}}}{a^{\frac{3 }{2}}}$
$\displaystyle =a^{\frac{3}{4}}+a^{\frac{3}{4}}+3a^{\frac{3}{4}}= 5a^{\frac{3}{4}}}=5\sqrt[3]{a^{4}}$
2)
$\displaystyle (2\sqrt{7}-3)\cdot (2\sqrt{7}+3)-(\sqrt{7}+1)^2-(\sqrt{7}-2)^2$
$\displaystyle [(2\sqrt{7})^2-9]-(7+2\sqrt{7}+1}-(7-4\sqrt{7}+4)=28-9-7-2\sqrt{7}-1-7+4\sqrt{7}-4=2\sqrt{7}$